# Quod Erat Demonstrandum

## 2011/03/31

### To 6C students

Filed under: Pure Mathematics — johnmayhk @ 5:23 上午

Just a sketch of the solutions to 1-mark questions in the previous quiz.

Q.7

Denote [$x$] as the greatest integer not greater than $x$.

Does $\displaystyle \lim_{x\rightarrow 0-}(\frac{x}{2011}[\frac{2012}{x}]-\frac{2011}{x}[\frac{x}{2012}])$ exist?

Hint:

$x$ is negative $\Rightarrow \frac{2012}{x}$ is negative.

Also, as $x$ tends to 0-, $\frac{2012}{x}$ tends to negative infinity.

Let $\frac{2012}{x} = -N+r$, where $N$ is a non-negative integer and $0\le r<1$

Hence, $\frac{x}{2011}[\frac{2012}{x}]=\frac{1}{2011}\frac{2012}{-N+r}[-N+r]=\frac{1}{2011}\frac{2012}{-N+r}(-N)=\frac{2012}{2011}(\frac{1}{1-\frac{r}{N}})$

Thus, $\displaystyle \lim_{x\rightarrow 0-}\frac{x}{2011}[\frac{2012}{x}]$ exists.

Now, for $x$ tends to 0-, let $\frac{x}{2012}=-s$ where $0

$\frac{2011}{x}[\frac{x}{2012}]=\frac{2011}{-2012s}[-s]=\frac{2011}{2012s}$

Thus, when $x$ tends to 0-, the limit above does not exist.

Q.8

Establish
$\pi =2\times\frac{2}{\sqrt{2}}\times\frac{2}{\sqrt{2+\sqrt{2}}}\times\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\times\dots$

Hint:

$\sin x$
$\equiv 2\cos(\frac{x}{2})\sin(\frac{x}{2})$
$\equiv 2^2\cos(\frac{x}{2})\cos(\frac{x}{2^2})\sin(\frac{x}{2^2})$
$\dots$
$\equiv 2^n\cos(\frac{x}{2})\cos(\frac{x}{2^2})\dots \cos(\frac{x}{2^n})\sin(\frac{x}{2^n})$

Hence

$\frac{\sin x}{2^n\sin(\frac{x}{2^n})}\equiv \cos(\frac{x}{2})\cos(\frac{x}{2^2})\dots \cos(\frac{x}{2^n})$

(for $\sin(\frac{x}{2^n}))\ne 0$)

Taking $n$ tends to infinity, yield

$\frac{\sin x}{x}=\cos(\frac{x}{2})\cos(\frac{x}{2^2})\cos(\frac{x}{2^3})\dots$

Put $x=\frac{\pi}{2}$, yield

$\frac{2}{\pi}=\cos(\frac{\pi}{2^2})\cos(\frac{\pi}{2^3})\cos(\frac{\pi}{2^4})\dots$

Now,

$\cos(\frac{\pi}{2^2})=\frac{1}{\sqrt{2}}$

$\cos(\frac{\pi}{2^3})=\sqrt{\frac{1+\cos(\frac{\pi}{2^2})}{2}}=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}}=\frac{\sqrt{2+\sqrt{2}}}{2}$

$\cos(\frac{\pi}{2^4})=\sqrt{\frac{1+\frac{\sqrt{2+\sqrt{2}}}{2}}{2}}=\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}$

and so on.

## 3 則迴響 »

1. 老師您好，請問哪裏有像　Lim 或　dy/dx的練習呢？

迴響 由 andy wong — 2011/04/01 @ 8:36 下午 | 回應

• In web or in my blog? For latter, there should be some in the download page.

迴響 由 johnmayhk — 2011/04/01 @ 9:54 下午 | 回應

2. o…兩個都可以，謝謝

迴響 由 andy wong — 2011/04/01 @ 10:05 下午 | 回應